A man starts rowing his stationary cuboidal boat of base area `A=10m^(2)`. The driving force on the boat due to rowing is `100N` in the direction of motion. Find the maximum velocity that the boat can achieve. Also find the time in which he will attain half of this maximum velocity. `[` Take coefficient of viscosity of water `=15` poise `]` The depth of the lake is `10m` and the combined mass of man and the boat to be `150kg. (u=0`, velocity gradient uniform)

Explanation

To solve the problem step by step, we will first find the maximum velocity that the boat can achieve and then calculate the time taken to reach half of this maximum velocity.

### Step 1: Understand the Forces Acting on the Boat

When the man rows the boat, a driving force \( F_{\text{external}} = 100 \, \text{N} \) is applied in the direction of motion. The boat experiences a viscous drag force \( F_{\text{viscous}} \) due to the viscosity of water, which opposes the motion. At maximum velocity, these two forces will balance each other.

### Step 2: Write the Equation for Maximum Velocity

The viscous force can be expressed using the formula:
\[ F_{\text{viscous}} = \eta \cdot A \cdot \frac{V_{\text{max}}}{D} \]
where:
- \( \eta = 15 \, \text{poise} = 15 \times 0.1 \, \text{N s/m}^2 = 1.5 \, \text{N s/m}^2 \) (conversion from poise to SI units),
- \( A = 10 \, \text{m}^2 \) (base area of the boat),
- \( D = 10 \, \text{m} \) (depth of the lake),
- \( V_{\text{max}} \) is the maximum velocity we want to find.

At maximum velocity, we have:
\[ F_{\text{external}} = F_{\text{viscous}} \]
Substituting the values, we get:
\[ 100 = 1.5 \cdot 10 \cdot \frac{V_{\text{max}}}{10} \]

### Step 3: Simplify and Solve for Maximum Velocity

Now, simplify the equation:
\[ 100 = 1.5 \cdot V_{\text{max}} \]
\[ V_{\text{max}} = \frac{100}{1.5} = \frac{200}{3} \, \text{m/s} \approx 66.67 \, \text{m/s} \]

### Step 4: Find Time to Reach Half of Maximum Velocity

To find the time taken to reach half of the maximum velocity \( V_{\text{max}}/2 \), we need to set up the equation of motion.

1. The net force acting on the boat at any time \( t \) is given by:
\[ F_{\text{net}} = F_{\text{external}} - F_{\text{viscous}} = m \cdot a \]
where \( a = \frac{dV}{dt} \).

2. The viscous force at any velocity \( V \) is:
\[ F_{\text{viscous}} = 1.5 \cdot 10 \cdot \frac{V}{10} = 1.5 V \]

3. Thus, the equation becomes:
\[ 100 - 1.5 V = 150 \frac{dV}{dt} \]

### Step 5: Rearranging and Separating Variables

Rearranging gives:
\[ \frac{dV}{100 - 1.5 V} = \frac{1}{150} dt \]

### Step 6: Integrate Both Sides

Integrate both sides from \( V = 0 \) to \( V = \frac{V_{\text{max}}}{2} = \frac{100/3}{2} = \frac{50}{3} \) and from \( t = 0 \) to \( t = t \):
\[ \int_0^{\frac{50}{3}} \frac{dV}{100 - 1.5 V} = \frac{1}{150} \int_0^t dt \]

The left side integrates to:
\[ -\frac{1}{1.5} \ln |100 - 1.5 V| \bigg|_0^{\frac{50}{3}} = -\frac{2}{3} \left( \ln(100 - 1.5 \cdot \frac{50}{3}) - \ln(100) \right) \]
Calculating gives:
\[ = -\frac{2}{3} \left( \ln(50) - \ln(100) \right) = -\frac{2}{3} \ln(0.5) \]

The right side gives:
\[ \frac{t}{150} \]

### Step 7: Solve for Time \( t \)

Equating both sides:
\[ -\frac{2}{3} \ln(0.5) = \frac{t}{150} \]
Thus,
\[ t = -100 \ln(0.5) \]

### Final Result

Calculating \( t \):
\[ t \approx 100 \cdot 0.693 \approx 69.3 \, \text{s} \]

### Summary of Results
- Maximum Velocity \( V_{\text{max}} = \frac{200}{3} \, \text{m/s} \approx 66.67 \, \text{m/s} \)
- Time to reach half of maximum velocity \( t \approx 69.3 \, \text{s} \)